Physics · Waves · NEET
The ends are tied down, so they cannot move at all. A point that never moves is a node (zero displacement). Since both ends are clamped, both ends are forced to be nodes. Every allowed vibration pattern must satisfy this: node at x = 0 and node at x = L. This boundary condition is what selects the allowed frequencies.
The distance between two neighbouring nodes is half a wavelength (lambda/2), not a full wavelength. A string fixed at both ends has a node at each end, and the string fits a whole number of these node-to-node gaps. So L must be a whole number of half-wavelengths: L = n(lambda/2). Rearranging gives lambda = 2L/n.
All harmonics. Because n can be any whole number 1, 2, 3, 4..., you get f_1, 2f_1, 3f_1, 4f_1 and so on. Odd-only harmonics belong to a pipe closed at one end (one node + one antinode). Do not mix them up: string fixed both ends = all harmonics; closed pipe = odd harmonics only.
Start with lambda = 2L/n. Use v = f(lambda), so f = v/lambda = v / (2L/n) = nv/2L. Therefore f_n = n(v/2L). The lowest (n = 1) is the fundamental f_1 = v/2L, and every higher frequency is a whole-number multiple of it.
In the nth harmonic the string vibrates in n loops (antinodes). So there are n antinodes and (n + 1) nodes (the two fixed ends plus the ones in between). Example: 3rd harmonic has 3 antinodes and 4 nodes.
Yes. For a string fixed at both ends, the harmonic number n equals the number of loops (antinodes) you see. So n = 2 means the second harmonic with 2 loops. This clean match happens because both ends are nodes.
Try the real previous-year questions from this chapter — each with the answer and a full solution.
It is the lowest allowed frequency, f_1 = v/2L, where v is the wave speed on the string and L is its length. Since v = sqrt(T/mu), we can also write f_1 = (1/2L)sqrt(T/mu).
f_n = n(v/2L) = (n/2L)sqrt(T/mu), where n = 1, 2, 3..., T is the tension, and mu is the linear mass density (mass per unit length).
lambda_n = 2L/n. The fundamental has lambda_1 = 2L (one loop), the second harmonic has lambda_2 = L, and so on.
The fixed ends set a boundary condition (nodes at both ends). Only frequencies that place a node exactly at each end survive; all others cancel out. This is why the string has a discrete set of natural (resonant) frequencies.
Both give all harmonics with f_n = n(v/2L), but the pipe has antinodes at both open ends while the string has nodes at both fixed ends. The pattern is flipped, yet the frequency formula is the same.
